This post is about a remarkable projective property of certain Boltzmann-Gibbs measures.

The model. Let $d\geq1$ and $g:\mathbb{R}^d\to(-\infty,+\infty]$ be continuous with $g(x)<\infty$ for all $x\neq0$. For all $\beta>0$, $n\geq2$, let $P_n$ be the probability measure on $(\mathbb{R}^d)^n$ with density proportional to $$(x_1,\ldots,x_n)\in(\mathbb{R}^d)^n\mapsto\exp(-\beta H(x_1,\ldots,x_n))$$ where $$H(x_1,\ldots,x_n)=\sum_{i=1}^n\frac{1}{2}|x_i|^2+\sum_{i\neq j}g(x_i-x_j).$$ Let $$X=(X_{n,1},\ldots,X_{n,n})\sim P_n.$$

A projection. Let $p:\mathbb{R}^d\to\mathbb{R}^d$ be an orthogonal projection on a subspace $E\subset\mathbb{R}^d$. Let $\pi$ and $\pi^\perp$ be the orthogonal projections on the subspaces $$L=\{(p(z),\ldots,p(z))\in(\mathbb{R}^d)^n:z\in\mathbb{R}^d\}\quad\text{and}\quad L^\perp.$$ We have, for all $x\in(\mathbb{R}^d)^n$,

$$\pi(x)=(p(s(x)),\ldots,p(s(x)))\quad\text{where}\quad s(x):=\frac{x_1+\cdots+x_n}{n}\in\mathbb{R}^d.$$

Indeed, we have
$$L^\perp=\{(x_1,\ldots,x_n)\in(\mathbb{R}^d)^n:(x_1+\cdots+x_n)\cdot p(z)=0\text{ for all }z\in\mathbb{R}^d\},$$
and for all $x=(x_1,\ldots,x_n)\in(\mathbb{R}^d)^n$ and all $(z,\ldots,z)\in L$, $z\in E$, we have
\begin{align}((x_1,\ldots,x_n)-(p(s(x)),\ldots,p(s(x))))\cdot(z,\ldots,z) &=\sum_{i=1}^nx_i\cdot z-np(s(x))\cdot z\\&=\sum_{i=1}^nx_i\cdot z-\sum_{i=1}^np(x_i)\cdot z\\&=\sum_{i=1}^n(x_i-p(x_i))\cdot z\\&=\sum_{i=1}^n0=0.\end{align}

Example 1: if $E=\mathbb{R}^d$ then $p(x)=x$.

Example 2 : $E=\mathbb{R}z$ for $z\in\mathbb{R}^d$ with $|z|=1$. Then $p(x)=(x\cdot z)z$.​​

The statement.  If all the ingredients are as above, then:

• $\pi(X)$ and $\pi^\perp(X)$ are independent random vectors;
• $\pi(X)$ is Gaussian with law $\mathcal{N}(0,\frac{1}{\beta}I_{\mathrm{dim}(E)})$ in an orthonormal basis of $L$
• $\pi^\perp(X)$ has law of density proportional to $x\in L^\perp\mapsto\mathrm{e}^{-\beta H(x)}$ with respect to the trace of the Lebesgue measure on the linear subspace $L^\perp$ of $(\mathbb{R}^d)^{n-1}$.

A proof. For all $x\in(\mathbb{R}^d)^n$, from $x=\pi(x)+\pi^\perp(x)$ we get
$$|x|^2=|\pi(x)|^2+|\pi^\perp(x)|^2$$
and on the other hand, for all $i,j\in{1,\ldots,n}$,

\begin{align}x_i-x_j&=\pi(x)i+\pi^\perp(x)_i-\pi(x)_j-\pi^\perp(x)_j\\&=s(x)+\pi^\perp(x)_i-s(x)-\pi^\perp(x)_j\\&=\pi^\perp(x)_i-\pi^\perp(x)_j.\end{align}

Since $V(x)=|x|^2$, it follows that for all $x=(x_1,\ldots,x_n)\in(\mathbb{R}^d)^n$, $$H(x)=|x|^2+\sum_{i\neq j}W(x_i-x_j)=|\pi(x)|^2+H(\pi^\perp(x)).$$

Let $u_1,\ldots,u_{dn}$ be an orthogonal basis of $(\mathbb{R}^d)^n=\mathbb{R}^{dn}$ such that $u_1,\ldots,u_{\mathrm{dim}(E)}$ is an orthonormal basis of $L$. For all $x\in(\mathbb{R}^d)^n$ we write $x=\sum_{i=1}^{dn}t_i(x)u_i$. We have $$\pi(x)=\sum_{i=1}^{\mathrm{dim}(E)}t_i(x)u_i\quad\text{and}\quad\pi^\perp(x)=\sum_{i=d+1}^{dn}t_i(x)u_i.$$ For all bounded measurable $f:L\to\mathbb{R}$ and $g:L^\perp\to\mathbb{R}$,
\begin{align}\mathbb{E}(f(\pi(X))g(\pi^\perp(X))) &=Z^{-1}\int_{(\mathbb{R}^d)^n}f(\pi(x))g(\pi^\perp(x))\mathrm{e}^{-\beta|\pi(x)|^2}\mathrm{e}^{-\beta H(\pi^\perp(x))}\mathrm{d}x_1\cdots\mathrm{d}x_n\\&=Z^{-1}\Bigr(\int_{\mathbb{R}^{\mathrm{dim}(E)}}f(t’)\mathrm{e}^{-\beta|t’|^2}\mathrm{d}t’\Bigr) \Bigr(\int_{\mathbb{R}^{d(n-1)}}g(t”) \mathrm{e}^{-\beta H(t”)}\mathrm{d}t”\Bigr)\end{align}

where $t’:=\sum_{i=1}^{\mathrm{dim}(E)}t_iu_i$, $\mathrm{d}t’:=\prod_{i=1}^{\mathrm{dim}(E)}\mathrm{d}t_i$, $t”:=\sum_{i=d+1}^{dn}t_iu_i$,$\mathrm{d}t”:=\prod_{i=d+1}^{dn}\mathrm{d}t_i$.